Question

How do you find the sum of the infinite geometric series given `Sigma(1.5)(0.25)^(n-1)` from n=1 to `oo`?

Answer 1

`Sigma_(n->oo)(1.5)(0.25)^(n-1)=2`

Explanation:

As in expanded form this series `Sigma(1.5)(0.25)^(n-1)` is

`1.5+(1.5)(0.25)+(1.5)(0.25)^2+(1.5)(0.25)^3+(1.5)(0.25)^4+ .......`,

it is a geometric series whose first term `a_1` is `1.5` and common ratio `r` is `0.25`.

As `r<1`, the sum of infinite series is `a/(1-r)=1.5/(1-0.25)=1.5/0.75=2`

Hence `Sigma_(n->oo)(1.5)(0.25)^(n-1)=2`